By [1]}, \(M\)  is bounded on \(\Phi = \mathbb {E}_w\) . By the inverse Hölder inequality, there exist \(p\)  and \(q\)  with \(1<p<p_{\mathbb {E}}\le q_{\mathbb {E}} <q <\infty \)  such that \(w\in A_p\subset A_q\) . Therefore, by Muckenhoupt's theorem, \(M\)  is also bounded on \(L^p_w\)  and \(L^q_w\) .
One can easily check that the operator \(S(f) := M(f w)/w\)  \((f\in L^1_{loc})\)  is bounded on
\( L^{p^{\prime }}_{w}\)  and \( L^{q^{\prime }}_{w}\) . Hence, it is of joint weak type \((q^{\prime },q^{\prime };p^{\prime },p^{\prime })\)  with respect to \((\mathbb {R}, w\mathrm {d}t)\) ; see [2]}. Consequently,
since \(p_{\mathbb {E}^{\prime }} = q^{\prime }_{\mathbb {E}}\)  and \(q_{\mathbb {E}^{\prime }}= p^{\prime }_\mathbb {E}\) , we have that \(1<q^{\prime }<p_{\mathbb {E}^{\prime }}\le q_{\mathbb {E}^{\prime }}< p^{\prime }<\infty \) . Next, by Boyd's theorem [3]}, we get that \(S\)  is bounded on \((\mathbb {E}^{\prime })_w\) . Moreover, since \((\mathbb {R}, w\mathrm {d}t)\)  is a resonant space (see [3]}, we have \((\mathbb {E}^{\prime })_w = {(\mathbb {E}_w)^{\prime }}^w\)  with equal norms, where \({(\mathbb {E}_w)^{\prime }}^w\)  denotes the dual space of \(\mathbb {E}_w\)  with respect to \((\mathbb {R}, w\mathrm {d}t)\) . Indeed, the inclusion \((\mathbb {E}^{\prime })_w \subset {(\mathbb {E}_w)^{\prime }}^{w}\)  follows from [3]} and the reverse one from the Luxemburg representation theorem [3]}.
Finally, note that for every \(f\in (\mathbb {E}_w)^{\prime }\)
\(\Vert Mf\Vert _{(\mathbb {E}_w)^{\prime }} & = \Vert (Mf)w^{-1}\Vert _{{(\mathbb {E}_w)^{\prime }}^w} = \Vert  S(fw^{-1})\Vert _{(\mathbb {E}^{\prime })_w}\\& \le \Vert S\Vert \Vert fw^{-1}\Vert _{{(\mathbb {E}^{\prime })_w}} \\& \le \Vert S\Vert  \Vert f\Vert _{{(\mathbb {E}_w)^{\prime }}},\) 

where \(\Vert S\Vert \)  denotes the norm of \(S\)  on \((\mathbb {E}^{\prime })_w\) . This completes the proof.
